📚 Permutations and Combinations | 排列与组合
Permutations and combinations are fundamental counting tools that allow us to determine the number of ways objects can be arranged or selected without having to list all possibilities. In IB Mathematics, mastering these concepts is essential for solving probability problems and understanding binomial expansions. This article breaks down the core principles, key formulas, common pitfalls, and applications, providing a complete revision guide.
排列与组合是基本的计数工具，让我们无需列出所有可能就能求出物体排列或选择的方法总数。在IB数学中，掌握这些概念对解决概率问题、理解二项式展开至关重要。本文拆解核心原理、关键公式、常见易错点和应用，提供一份完整的复习指南。
1. The Fundamental Counting Principle | 基本计数原理
ur in m × n ways. This multiplication principle extends to any number of sequential tasks.
如果一件事有m种不同的发生方式，随后另一件事有n种不同的发生方式，那么这两件事相继发生共有m × n种方式。这个乘法原理可推广到任意多个连续的任务。
osing a starter from 3 soups and a main course from 5 dishes gives 3 × 5 = 15 possible two‑ ces are independent: if you have 4 shirts and 3 trousers, you can form 4 × 3 = 12 outfits.
例如，从3种汤中选一道前菜，从5道主菜中选一道，共有3 × 5 = 15种两菜套餐。选择相互独立时同样适用：若有4件衬衫和3条裤子，可搭配出4 × 3 = 12套着装。
When tasks are mutually exclusive—meaning you can do one or the other but not both—the addition principle velling from city A to B by either 2 flights or 3 trains yields 2 + 3 = 5 distinct routes.
当事务互斥——即只能做其中一件而不可兼得——则使用加法原理：总方式数 = m + n。例如，从A城到B城可乘2个航班或3趟火车，共有2 + 3 = 5条不同路线。
2. Factorial Notation | 阶乘记号
For a positive integer n, n factorial, written as n!, is the product of all integers from n down to 1: n! = n × (n − 1) × (n − 2) × … × 3 × 2 × 1. By definition, 0! = 1. Factorials grow rapidly; 5! = 120, 10! = 3,628,800.
对于正整数n，n的阶乘记作n!，表示从n递减到1的所有整数之积：n! = n × (n − 1) × (n − 2) × … × 3 × 2 × 1。定义0! = 1。阶乘增长极快；5! = 120，10! = 3,628,800。
l distinct objects. The number of ways to arrange n different items in a line is exactly n!. For three letters A, B, C, the 3! = 6 arrangements are ABC, ACB, BAC, BCA, CAB, CBA.
阶乘能高效计算所有不同物体的排列数。将n个不同物体排成一列的排列数正好是n!。对三个字母A、B、C，3! = 6种排列为ABC、ACB、BAC、BCA、CAB、CBA。
nter factorial expressions in permutation and combination formulas. Simplify ratios like n!/(n−2)! by cancelling common factors: n!/(n−2)! = n × (n−1). Cancelling is a key skill in algebraic manipulation of these expressions.
在排列组合公式中常遇到阶乘表达式。通过约分可简化如n!/(n−2)!：n!/(n−2)! = n × (n−1)。约分是处理此类代数式的关键技巧。
3. Permutations (Linear Arrangements) | 排列（线性排列）
A permutation is an ordered arrangement of objects. The number of ways to arrange r objects chosen from n distinct objects is denoted by P(n, r) or ⁿPᵣ and is calculated using the formula:
排列是指对象的有序排列。从n个不同物体中选出r个进行排列的方法数记作P(n, r)或ⁿPᵣ，公式为：
P(n, r) = n! / (n − r)!
This is derived by considering n choices for the first position, (n−1) for the second, …, down to (n−r+1) for the r‑th position. Multiplying gives n × (n−1) × … × (n−r+1), which equals n!/(n−r)!.
推导思路是：第一个位置有n种选择，第二个有(n−1)种，……，第r个位置有(n−r+1)种。相乘得到n × (n−1) × … × (n−r+1)，即n!/(n−r)!。
ays to award gold, silver and bronze medals among 8 runners (order matters) is P(8, 3) = 8! / 5! = 8 × 7 × 6 = 336.
例如，从8名选手中颁发金、银、铜牌（顺序重要）的方法数为P(8, 3) = 8! / 5! = 8 × 7 × 6 = 336。
When all n objects are arranged (r = n), the formula reduces to P(n, n) = n! / 0! = n!, confirming that arranging all distinct items in a line yields n! permutations.
当排列所有n个物体时（r = n），公式简化为P(n, n) = n! / 0! = n!，确认将所有不同物体排成一列共有n!种排列。
4. Permutations with Repetition | 有重复的排列
When some objects are identical, simply applying n! overcounts arrangements because swapping identical items does not produce a new arrangement. If there are n items in total, with r₁ of one type, r₂ of another, …, rₖ of the k‑th type, the number of distinct permutations is:
当存在相同物体时，直接使用n!会重复计数，因为交换相同物体不会产生新排列。若共有n个物体，其中第1类有r₁个，第2类有r₂个，……，第k类有rₖ个，则不同排列数为：
n! / (r₁! × r₂! × … × rₖ!)
For example, the word 'MISSISSIPPI' has 11 letters: 1 M, 4 I's, 4 S's, 2 P's. The number of distinct arrangements is 11! / (1! × 4! × 4! × 2!) = 34,650.
例如，单词“MISSISSIPPI”有11个字母：1个M，4个I，4个S，2个P。不同排列数为11! / (1! × 4! × 4! × 2!) = 34,650。
This formula is obtained by first pretending all items are distinct (giving n! arrangements) and then dividing by the factorial of the frequency of each identical group to eliminate overcounting from their internal swaps.
该公式通过先假设所有物体不同（得n!），再除以每一组相同物体内部交换导致的重复计数（各组频数的阶乘）来消除重复。
Be careful: the denominator includes a factorial for every class of repeated items, including those that appear only once (1! = 1, so it does not change the value).
注意：分母要对每一类重复物体取阶乘，即使只出现一次也要写1!（因为1! = 1，不影响结果），以保持思维清晰。
5. Combinations | 组合
A combination is a selection of objects where order does not matter. The number of ways to choose r objects from n distinct objects is denoted by C(n, r), ⁿCᵣ or ⎛n⎞⎝r⎠, and is given by:
组合是指顺序不重要的选择方式。从n个不同物体中选出r个的组合数记作C(n, r)、ⁿCᵣ或⎛n⎞⎝r⎠，公式为：
C(n, r) = n! / [r! × (n − r)!]
This comes from the permutation count P(n, r) and then dividing by r! because each set of r items can be arranged in r! ways but all represent the same combination when order is ignored.
该式由排列数P(n, r)除以r!得出：每r个物体的集合可以有r!种排列，但在忽略顺序时它们对应同一个组合。
For example, choosing a committee of 3 people from 10 candidates yields C(10, 3) = 10! / (3! × 7! d posts (like president, secretary, treasurer), it would be a permutation: P(10, 3) = 720.
例如，从10名候选人中选出3人组成委员会，有C(10, 3) = 10! / (3! × 7!) = 120种方式。若委员会有职务分工（主席、秘书、财务），则属于排列：P(10, 3) = 720。
). Choosing 2 items from 5 is the same as choosing 3 to leave out: C(5, 2) = C(5, 3) = 10.
一个重要恒等式是C(n, r) = C(n, n−r)。从5个中选2个等同于选择留下3个：C(5, 2) = C(5, 3) = 10。
6. Relationship Between Permutations and Combinations | 排列与组合的关系
The connection P(n, r) = C(n, r) × r! rst choosing the objects (combination) and then arranging them. This is a powerful problem‑solving strategy.
基本关系式P(n, r) = C(n, r) × r!至关重要。它说明排列计数可先选物体（组合）再将其进行排列。这是一种强大的解题策略。
r others. For instance, selecting a team of 5 from 12 and then choosing a captain and vice‑captain from the 5 involves a combination followed by a permutation: C(12, 5) × P(5, 2).
许多IB题目要求处理顺序在某些部分重要、某些部分不重要的情境。例如，从12人中选5人组队，再从5人中选队长和副队长，先组合后排列：C(12, 5) × P(5, 2)。
Distinguishing between "select" and "arrange" cues in the problem statement is essential. Words like "committee", "group", "set" suggest combination; "rank", "order", "sequence", "arrangement" suggest permutation.
在题目表述中区分“选择”和“排列”的关键词至关重要。如“委员会”、“小组”、“集合”等暗示组合；而“排名”、“顺序”、“序列”、“排列”则暗示排列。
7. Applications to Probability | 概率应用
bility P(A) = n(A) / n(S), where n(S) is the total number of outcomes in the sample space.
概率计算常归于使用组合或排列来计数有利结果数和所有可能结果数，前提是等可能事件。概率 P(A) = n(A) / n(S)，其中n(S)为样本空间的结果总数。
52, 2); favourable ways is C(4, 2). Probability = C(4, 2) / C(52, 2) = 6 / 1326 ≈ 0.00452.
例如，从一副52张标准扑克中无放回地抽出2张A：任意2张的组合总数为C(52, 2)；有利组合数为C(4, 2)。概率 = C(4, 2) / C(52, 2) = 6 / 1326 ≈ 0.00452。
When order matters in the experiment (e.g. drawing cards sequentially with care for the sequence), use permutations for both numerator and denominator consistently. The key is to be consistent: if the sample space is counted with order, favourable outcomes must be too.
若试验中顺序重要（如依次抽牌且关注顺序），则分子分母都需使用排列计数。核心是保持一致性：样本空间若考虑顺序，有利事件也须考虑顺序。
The hypergeometric distribution provides a model, but at this level, direct counting with combinations and the addition and multiplication principles is more instructive and builds intuition.
超几何分布提供了模型，但在当前阶段，直接借助组合、加法原理和乘法原理进行计数更有启发性，也助于建立直觉。
8. Arrangements with Restrictions | 带限制条件的排列
Restrictions such as "two specific people must sit together" or "a certain object cannot be placed at the ends" require strategic counting. Treat items that must be together as a single block, arrange the block and the remaining items, then arrange within the block.
条件如“某两人必须相邻而坐”或“某个物体不可置于两端”，需要策略性计数。将必须相邻的物体视为一个整体，先排列这个整体和其他物体，再在整体内部进行排列。
Example: In how many ways can 5 boys and 3 girls line up if all girls must stand together? Treat the girls as one block, giving 6 objects (5 boys + 1 block) → 6! ways. Inside the block, the 3 girls can be arranged in 3! ways. Total = 6! × 3! = 720 × 6 = 4,320.
例题：5个男孩和3个女孩排队，要求所有女孩必须站在一起。把女孩视为一个整体，共6个物体（5个男孩+1个整体）→ 6!种排法。整体内部3个女孩有3!种排列。总数 = 6! × 3! = 720 × 6 = 4,320。
For "must not be at the ends" restrictions, fill the restricted positions first or use the complement rule: total arrangements minus those where the restriction is violated. Both approaches are valid and help develop flexible counting skills.
对于“不可在两端”的限制，可优先填充受限位置，或使用补集法：总排列数减去违反限制的排列数。两种方法均有效，有助于培养灵活的计数能力。
When multiple restrictions exist, breaking the problem into cases (using the addition principle) is often necessary. For example, "at least one end is occupied by a boy" could be solved by subtracting 'both ends are girls' from the total.
当存在多个限制时，常常需要分情况讨论（使用加法原理）。例如“至少一端是男孩”可通过从总数中减去“两端都是女孩”来求解。
9. Circular Permutations | 圆排列
When objects are arranged in a circle, rotations of the same ordering are considered identical because there is no fixed starting point. The number of ways to arrange n distinct objects around a circle is (n − 1)!. Essentially, fix one object to break the rotational symmetry, then arrange the remaining (n−1) objects linearly.
当物体排成圆圈时，旋转相同的排列视为同一种，因为没有固定起点。将n个不同物体排成一个圆圈的排列数为 (n − 1)!。方法是固定一个物体以打破旋转对称，再将剩下的(n−1)个物体进行线性排列。
For example, seating 6 people around a round table gives (6 − 1)! = 5! belled (e.g. numbered chairs), the rotation symmetry is broken and the answer reverts to 6! linear permutations.
例如，6人围坐一张圆桌，有 (6 − 1)! = 5! = 120 种坐法。如果椅子有编号，旋转对称被打破，答案恢复为6!线性排列。
When there are identical objects in a circle, the formula becomes more complex, but IB typically tests distinct objects. Be mindful if reflections are considered identical (e.g. a necklace can be flipped), halving the count again: (n − 1)! / 2 for n > 2.
当圆圈中有相同物体时，公式更为复杂，但IB通常考查不同物体。注意若考虑翻转对称（如项链可翻转），则需再除以2：对n > 2，环形排列数为 (n − 1)! / 2。
Always read the context: "arranged in a circle" implies rotational symmetry; "arranged around a table with distinct chairs" removes that symmetry and uses standard linear permutations.
务必审题：“围成一圈”隐含旋转对称；“坐在有编号的椅子上围桌”则消除了对称性，仍用线性排列。
10. Combinations with Repetition | 可重复组合
objects from n types, with repetition permitted, is C(r + n − 1, r) = C(r + n − 1, n − 1).
当允许重复选择物体且顺序不重要时，使用可重复组合（也称星条法）。从n种类型中选出r个物体（可重复）的方法数为 C(r + n − 1, r) = C(r + n − 1, n − 1)。
5) = 21. This counts the ways to distribute the 5 identical selections among 3 categories.
例题：从3种甜甜圈（原味、巧克力、果酱）中选5个，允许重复。此时 n = 3, r = 5。可能组合数为 C(5+3−1, 5) = C(7, 5) = 21。这计算的是将5个相同选择分配到3个种类中的方式。
he counting principle extension. It links to integer solutions of equations like x₁ + x₂ + … + xₙ = r, where xᵢ are non‑negative integers.
此主题在某些IB大纲中出现，特别是计数原理的拓展。它与方程 x₁ + x₂ + … + xₙ = r 的非负整数解个数相关联。
11. Pascal’s Triangle and Binomial Coefficients | 帕斯卡三角形与二项式系数
Combinations C(n, r) appear as entries in Pascal' n r gives C(n, r). The triangle reveals identities like C(n, r) = C(n−1, r−1) + C(n−1, r).
组合数 C(n, r) 出现在帕斯卡三角形中，每个数是其上方两数之和。第n行对应n，第r个位置给出C(n, r)。该三角形揭示了如 C(n, r) = C(n−1, r−1) + C(n−1, r) 的恒等式。
Pascal’s triangle also provides the coefficients for the binomial expansion (a + b)ⁿ. The coefficient of aⁿ⁻ʳ bʳ is C(n, r). This is why C(n, r) are called binomial coefficients.
帕斯卡三角形同时给出二项式展开 (a + b)ⁿ 的系数。项 aⁿ⁻ʳ bʳ 的系数正是 C(n, r)，因此C(n, r) 也称二项式系数。
of binomial expansions deepens your grasp of both topics: expanding (1 + x)ⁿ = Σ C(n, r) xʳ is justified by choosing r 'x's from n factors.
理解二项式展开的组合意义能加深对两个主题的掌握：(1 + x)ⁿ = Σ C(n, r) xʳ 的推导正是从n个因式中选出r个x的组合方式。
12. Summary and Common Pitfalls | 总结与常见错误
Always ask: "Does order matter?" If yes, use permutations; if no, use combinations. Count systematically by breaking tasks into stages, applying the multiplication or addition principles appropriately. Identify identical objects and divide by the relevant factorials to correct overcounting.
始终自问：“顺序重要吗？”若是，用排列；若否，用组合。系统计数，将任务分解为阶段，适当运用乘法或加法原理。识别相同物体，并除以相应阶乘以修正重复计数。
Common mistakes: confusing C(n, r) with P(n, r); forgetting to divide by r! in combinations; mishandling identical items; using permutations when the problem asks for unordered selections; ignoring the effect of fixed positions in circular arrangements; and applying the addition principle when events are not mutually exclusive.
常见错误：混淆 C(n, r) 与 P(n, r)；组合时忘记除以 r!；相同物体处理错误；题目要求无序选择时用了排列；环形排列时忽略固定位置的影响；事件不互斥时错用加法原理。
Practice with a variety of problems—from basic arrangements to multi‑stage probability questions—to build fluency. Check your answers with small numbers by listing if possible, and use calculator functions nPr and nCr only after you are confident in setting up the expression correctly.
通过多种问题练习——从基础排列到多阶段概率题——来培养熟练度。对于小数目可尝试列举验证答案；只有在确信正确列出表达式后，再使用计算器的 nPr 和 nCr 功能。
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